(a) Find the indefinite integrals of the following functions.

(i) f (t) = 6 cos(3t) + 5e^−10t

(ii) g(x) = 21-12x^3/ x (x > 0)

(iii) h(u) = cos^2( 1/8 u)

(b) Evaluate: (this big F sign at the start, 5 at the top and 1 at the bottom)
5 1/4x (7 + 6x^2) dx
F
1

(c)
(i) Write down a definite integral that will give the value of the area under the curve y = x^2 cos(2x) between x = 3/4 pie and x = pie.

(You are not asked to evaluate the integral by hand.)

(ii) find the area described in part (c)(i), giving your
answer correct to three decimal places

MY ANSWERS:
for a) ii)

6cos (3t) dx = -4sin (3t) + c
5e^10t dx = 1/5 e^-10t + c

ii) g (x) = 21- 12x^3 / x (x>0)

diverse through by x

g (x) = 21/ x - 12x^3

21/ x - 12x^3 dx = 12Ln x - x^4 + c

iii) h (u) = cos^2 (1/8u)

using the second version of the double angled formula for cos (2delta)

=1/2 cos (2x 1/8 u) + 1/2 du
= 1/2 / 28 sin (2/8u) + 1/2 x + c

= 6/4 sin (2/8u) + 1/2u + c

i don't know whether that's right, and i don't know how to do the rest.

could some body please help me out. thanks!

For part (a), your answers for (i) and (ii) are correct. However, the answer for (iii) is not correct.

(a)

(i) The indefinite integral of f(t) = 6 cos(3t) + 5e^(-10t) is:
∫[6 cos(3t) + 5e^(-10t)] dt = 2 sin(3t) - (1/2)e^(-10t) + C

(ii) The indefinite integral of g(x) = (21 - 12x^3)/x (for x > 0) is:
∫[(21/x) - 12x^2] dx = 21 ln(x) - 4x^3 + C

(iii) The indefinite integral of h(u) = cos^2((1/8)u) is:
∫cos^2((1/8)u) du = (1/2)(u + sin((1/4)u)) + C

For part (b), the integral is written as ∫[5/(4x)](7 + 6x^2) dx, with the limits of integration from x = 1 to x = 5. The correct answer would be:

∫[5/(4x)](7 + 6x^2) dx = 5/4 ∫[(7 + 6x^2)/x] dx
= 5/4 ∫[7/x + 6x] dx
= 5/4 [7 ln|x| + 3x^2] + C

To evaluate this integral from x = 1 to x = 5, you would substitute the limits of integration into the expression and subtract the lower limit from the upper limit.

∫[5/(4x)](7 + 6x^2) dx [from x = 1 to x = 5]
= (5/4)[7 ln|5| + 3(5)^2] - (5/4)[7 ln|1| + 3(1)^2]
= (5/4)[7 ln(5) + 75 - 7 ln(1) - 3]
= (5/4)(7 ln(5) + 72)

For part (c):

(i) The definite integral that gives the value of the area under the curve y = x^2 cos(2x) between x = 3/4π and x = π is:
∫[x^2 cos(2x)] dx [from x = 3/4π to x = π]

(ii) To find the value of this definite integral, you would need to evaluate it numerically. Since you are not asked to do so, the exact value of the area under the curve cannot be determined with the given information.